Question

find the derivative of $y=sqrt{x}ln x$.

Explanation:

Step1: Apply product rule

The product rule is \((uv)^\prime = u^\prime v+uv^\prime\). Let \(u = \sqrt{x}=x^{\frac{1}{2}}\), \(v=\ln x\). First find \(u^\prime\) and \(v^\prime\).
\(u^\prime=\frac{1}{2}x^{\frac{1}{2}- 1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\), \(v^\prime=\frac{1}{x}\)

Step2: Substitute into product rule

\(y^\prime=u^\prime v + uv^\prime=\frac{1}{2\sqrt{x}}\cdot\ln x+\sqrt{x}\cdot\frac{1}{x}\)
Simplify the second - term \(\sqrt{x}\cdot\frac{1}{x}=x^{\frac{1}{2}}\cdot x^{-1}=x^{\frac{1}{2}-1}=x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}\)
So \(y^\prime=\frac{\ln x}{2\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{\ln x + 2}{2\sqrt{x}}\)

Answer:

\(\frac{\ln x + 2}{2\sqrt{x}}\)